Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.
D * x +(100- D) * 2x= 175x
=> D= 25 days
Now , the work done in 25 days = 25x
Total work = 175x
Therefore, workdone before increasing the no of workers = % =
Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2
Therefore, A's 1 day's work=1/9
B's 1 day's work=1/18
(A+B)'s 1 day's work= 1/9 + 1/18 = 1/6
so, A and B together can finish the work in 6 days
(A+B)'s 1 day's work=1/20
C's 1 day work=1/60
(A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15
Also A's 1 day's work =(B+C)'s 1 day's work
Therefore, we get: 2 x (A's 1 day 's work)=1/15
=>A's 1 day's work=1/30
Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60
So, B alone could do the work in 60 days.
2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8
=>(A+B+C)'s 1 day's work= 1/16
work done by A,B and C in 10 days=10/16 = 5/8
Remaining work= 3/8
A's 1 day's work=
Now, 1/48 work is done by A in 1 day.
So, 3/8 work wil be done by A in =48 x (3/8) = 18 days
P can complete the work in (12 x 8) hrs = 96 hrs
Q can complete the work in (8 x 10) hrs=80 hrs
Therefore, P's 1 hour work=1/96 and Q's 1 hour work= 1/80
(P+Q)'s 1 hour's work =(1/96) + (1/80) = 11/480. So both P and Q will finish the work in 480/11 hrs
Therefore, Number of days of 8 hours each = (480/11) x (1/8) = 60/11
1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192
Work done in 6 days=
Remaining work = 1/4
(8 men +8 women)'s 1 day work = =1/8
Remaining work =1/4 - 1/8 = 1/8
1/96 work is done in 1 day by 1 man
Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men
(A+B+C) do 1 work in 10 days.
So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5
Remaining work=1-2/5=3/5
(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50
Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25
A does 1/25 work in in 1 day
Therefore 1 work in 25 days.
A : C
Efficiency 5 : 3
No of days 3x : 5x
Given that, 5x-6 =3x => x = 3
Number of days taken by A = 9
Number of days taken by C = 15
B : C
Days 2 : 3
Therefore, Number of days taken by B = 10
Work done by B and C in initial 2 days = = 1/3
Thus, Rest work =2/3
Number of days required by A to finish 2/3 work = (2/3) x 9 = 6 days
Let the number of workers be x.
Now, Using work equivalence method,
X + (X-1) + (X-2)+ . . . . + 1 = X *55% of X
=> [X * (X+1)] / 2 = X * (55X/100) [because, Series is in AP. Sum of AP = {No. of terms (first term+ last term)/2} ]
Therefore, X = 10
=> x= 20 and x=-14
so, the acceptable values is x=20
Therefore, Total work =(x-2)x = 18 x 20 =360 unit
Now 360 = 30 x k
=> k=12 days
Let the two conditioners be A and B
'A' cools at 40min
'B' at 45min
Together = (a x b)/(a + b)
= (45 x 40)/(45 + 40)
= 45 x 40/85
= 21.1764
= 22 min (approx).
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